# isoperimetric inequality proof

While going through Pressley's Elementary Differential Geometry I came across Isoperimetric inequality.For a simple closed curve $\gamma$ with length $l$ and area of interior $A$ , $4 \pi A \leq l^{2}$. Equality holds if and only if the curve is a circle.

Author uses Wirtinger's inequality to prove the result.

Firstly he makes assumption that the curve is of parameter $t= \frac{\pi s}{l}$. $\gamma$ is a unit sped curve and having the period $\pi$. Also makes the assumption that curve begins and ends at origin.

The he starts with parametric equations $x=rcos \theta$ and $y= rsin \theta$ and proves the inequality. I cannot understand why he uses the the parametric equation of a circle for general curve.

If we proceed in this way the later part of the proof follows by routine calculations.

• What exactly are you asking? Is it just the fact that $r$ is not a constant, but a function of $\theta$ that was the problem? Is it something else? – robjohn Mar 2 '18 at 20:24

The keyword is polar coordinates. Given any curve $x:I\to\Bbb R^2\setminus\{(0,0)\}$, there are functions

$$r:I\to\Bbb R^+,\quad \theta:I\to\Bbb R$$

so that

$$x(t)=\begin{pmatrix}r(t)\cos\theta(t)\\r(t)\sin\theta(t)\end{pmatrix}.$$

So these values $r$ and $\theta$ given in these equations are no constants, but are functions of $t$ themselves. The result can be any curve (strictly speaken: any curve which does not pass through the origin, but we can solve this by continuity).

Example. It is a bit hard to write this down for the ellipse (as requested by you in a comment) because the ellipse is already hard to reparametrize to arc length (which is a requirement in your post). Therefore I will demonstrate it for the square with corners $(0,0), (1,0), (1,1)$ and $(0,1)$. An arc length parametrization $x:I\to\Bbb R^2$ must be defined on the interval $[0,4]$. We then have

$$r(t)=\begin{cases} t & \text{for t\in[0,1]} \\ \sqrt{t^2-2t+2} & \text{for t\in[1,2]} \\ \sqrt{t^2-6t+10} & \text{for t\in[2,3]} \\ 4-t & \text{for t\in[3,4]} \end{cases},\qquad \theta(t)=\begin{cases} 0 & \text{for t\in[0,1]} \\ \arctan(t-1) & \text{for t\in[1,2]} \\ \pi/2-\arctan(3-t) & \text{for t\in[2,3]} \\ \pi/2 & \text{for t\in[3,4]} \\ \end{cases}.$$

Example. Note that your post requires that the curve starts in $(0,0)$, hence a full circle is not parametrized by $r=\mathrm{const}$ (this would set the circles center at $(0,0)$ instead). But think about the unit circle around the point $(1,0)$. It can be parametrized on $I=[0,2\pi]$ by

$$r(t)=\sqrt{2+2\cos t},\qquad \theta(t)=\arctan\left(\frac{\sin(t)}{\cos(t)+1}\right).$$

Note: in both examples I dropped the requirement $I=[0,\pi]$ as given in your post, but I think this is no problem as you can simply scale the curves in question (this is what the author had in mind anyway).

$r$ is a function of $\theta$, so it is not just a circle.

• Can you explain ? Because I am still not clear as $x^{2}+y^{2}=1$ for this parametrisation. How can we write ellipse in the form $x=rcos \theta$ $y=rsin \theta$ ? what will be $r$ in the case of ellipse ? – Madhu Feb 17 '18 at 2:06
• This can only be achieved for star-shaped curves. In general, $\theta$ and $r$ must be functions of $t$. Especially, because the curve should be parametrized according to arc length, $r(\theta)$ is not always possible. – M. Winter Mar 2 '18 at 8:27

$$x^2+y^2=r(t)^2$$ the right-hand side of which is not constant but a function of $t$, so it does not describe a circle but an arbitrary closed curve.

On your secondary question: An ellipse can be written in various ways, for example if we are talking of an ellipse centred at the origin, $$(x,y)=(a\cos\theta, b\sin\theta), \text{ with } \theta\in[0,2\pi).$$ representing $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$ But, of course, this isn't the only possibility. If we represent this curve in polar coordinates: the distance $OM$ is $\sqrt{a^2\cos\theta^2 + b^2\sin\theta^2}$ and the angle between the $x$-axis an the point of interest is given by $$\cos \phi= \frac{a\cos\theta}{\sqrt{a^2\cos\theta^2 + b^2\sin\theta^2}},$$ and $$\sin \phi= \frac{b\sin\theta}{\sqrt{a^2\cos\theta^2 + b^2\sin\theta^2}}.$$ Provided that one can show that the map $\theta \to \phi$ is one to one, so that the map $\phi \to \theta$ is well defined, then your polar representation is $$r(\phi) = \sqrt{a^2\cos\theta(\phi)^2 + b^2\sin\theta(\phi)^2}.$$ Luckily, $\cos^{-1}$ and $\sin^{-1}$ have overlapping support, so you can check that $\phi^\prime(\theta)$ does not cancel, and therefore it is one to one.